Question

The eigenvalues of \( [A] = \begin{bmatrix} 2 & -3.5 & 6 \\ 3.5 & 5 & 2 \\ 8 & 1 & 8.5 \end{bmatrix} \) are
\( \lambda_1 = -1.547, \lambda_2 = 12.330 \), and \( \lambda_3 = 4.711 \).
The absolute value of the determinant of matrix A is ______ (rounded off to two decimal places).

Hint:

Use \( \det(A) = \lambda_1 \lambda_2 \lambda_3 \) for a 3x3 matrix, then take the absolute value.

Answer 1

Correct Answer: 89.86

Step 1: Recall the property connecting eigenvalues and determinant.
For any square matrix A, the determinant of A equals the product of all its eigenvalues, that is \( \det(A) = \lambda_1 \times \lambda_2 \times \lambda_3 \) for a 3x3 matrix. This is a standard property, it follows from the fact that the characteristic polynomial \( \det(A-\lambda I) = 0 \) can be written as \( (\lambda_1-\lambda)(\lambda_2-\lambda)(\lambda_3-\lambda) \), and putting \(\lambda=0\) gives \( \det(A) = \lambda_1\lambda_2\lambda_3 \).

Step 2: Multiply the first two eigenvalues.
\[ \lambda_1 \times \lambda_2 = (-1.547)(12.330) = -19.07451 \]

Step 3: Multiply the result by the third eigenvalue.
\[ \det(A) = -19.07451 \times 4.711 = -89.860 \]

Step 4: Take the absolute value.
The question asks for the absolute value of the determinant, so drop the negative sign.
\[ |\det(A)| = 89.86 \]

Final Answer:
The absolute value of the determinant of matrix A, rounded off to two decimal places, is 89.86.
\[ \boxed{89.86} \]